Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int kInf = 1e9;
int best = kInf;
vector<pair<int, int>> best_ans_row;
vector<pair<int, int>> best_ans_col;
void Process(int k, const vector<vector<int>>& matr) {
  int n = matr.size(), m = matr.begin()->size();
  int res = k;
  vector<int> h(n), v(m);
  h[0] = k;
  vector<pair<int, int>> ans_row = {make_pair(0, k)};
  vector<pair<int, int>> ans_col = {};
  for (int col = 0; col < m; ++col) {
    if (matr[0][col] - h[0] < 0) {
      return;
    }
    v[col] = matr[0][col] - h[0];
    res += v[col];
    ans_col.emplace_back(col, v[col]);
  }
  for (int row = 1; row < n; ++row) {
    h[row] = matr[row][0] - v[0];
    if (h[row] < 0) {
      return;
    }
    res += h[row];
    ans_row.emplace_back(row, h[row]);
    for (int col = 0; col < m; ++col) {
      if (matr[row][col] != h[row] + v[col]) {
        return;
      }
    }
  }
  if (res < best) {
    best = res;
    best_ans_row = ans_row;
    best_ans_col = ans_col;
  }
}
int main() {
  std::ios_base::sync_with_stdio(false);
  int n, m;
  cin >> n >> m;
  auto matr = vector<vector<int>>(n, vector<int>(m));
  for (int row = 0; row < n; ++row) {
    for (int col = 0; col < m; ++col) {
      cin >> matr[row][col];
    }
  }
  for (int k = 0; k <= matr[0][0]; ++k) {
    Process(k, matr);
  }
  if (best == kInf) {
    cout << -1 << '\n';
    return 0;
  }
  cout << best << '\n';
  for (auto& item : best_ans_row) {
    while (item.second--) {
      cout << "row " << item.first + 1 << '\n';
    }
  }
  for (auto& item : best_ans_col) {
    while (item.second--) {
      cout << "col " << item.first + 1 << '\n';
    }
  }
  return 0;
}